Triple integral over a simplex

Setup.
$$I = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} x \, dz \, dy \, dx.$$

Integration in $z$.
$$\int_0^{1-x-y} x \, dz = x(1-x-y).$$

Integration in $y$.
$$\int_0^{1-x} x(1-x-y) \, dy = x\left[ (1-x)y - \frac{y^2}{2} \right]_0^{1-x} = x \cdot \frac{(1-x)^2}{2}.$$

Integration in $x$.
$$I = \frac12 \int_0^1 x(1-x)^2 \, dx = \frac12 \int_0^1 (x - 2x^2 + x^3) \, dx = \frac12 \left( \frac12 - \frac23 + \frac14 \right) = \frac12 \cdot \frac{1}{12} = \frac{1}{24}.$$

Remark. The volume of the simplex is $1/6$, and the mean $x$-coordinate is $1/4$, consistent by symmetry.